题目:http://pat.zju.edu.cn/contests/pat-a-practise/1035

分析:简单题。直接搜索,然后替换,不会超时,但是应该有更好的办法。

题目描述:

To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) from O (o in uppercase). One solution is to replace 1 (one) by @, 0 (zero) by %, l by L, and O by o. Now it is your job to write a program to check the accounts generated by the judge, and to help the juge modify the confusing passwords.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N (<= 1000), followed by N lines of accounts. Each account consists of a user name and a password, both are strings of no more than 10 characters with no space.

Output Specification:

For each test case, first print the number M of accounts that have been modified, then print in the following M lines the modified accounts info, that is, the user names and the corresponding modified passwords. The accounts must be printed in the same order as they are read in. If no account is modified, print in one line "There are N accounts and no account is modified" where N is the total number of accounts. However, if N is one, you must print "There is 1 account and no account is modified" instead.

Sample Input 1:

3
Team000002 Rlsp0dfa
Team000003 perfectpwd
Team000001 R1spOdfa

Sample Output 1:

2
Team000002 RLsp%dfa
Team000001 R@spodfa

Sample Input 2:

1
team110 abcdefg332

Sample Output 2:

There is 1 account and no account is modified

Sample Input 3:

2
team110 abcdefg222
team220 abcdefg333

Sample Output 3:

There are 2 accounts and no account is modified

参考代码:

#include<iostream>
#include<string.h>
#include<string>
using namespace std; class User
{
public:
void set(string nam,string pas)
{
name = nam;
pass = pas;
is_Changed = false;
};
User(){};//此处要注意!!!如果写成User();则不能运行! void check();
void show();
bool is_Changed;
private:
string name;
string pass;
int len;
}; void User::show()
{
cout<<name<<" "<<pass<<endl;
} void User::check()
{
len = pass.length();
int i;
for(i=0; i<len; i++)
{
if(pass[i] == '1') {pass[i] = '@'; is_Changed = true;}
else if(pass[i] == '0') {pass[i] = '%'; is_Changed = true;}
else if(pass[i] == 'l') {pass[i] = 'L'; is_Changed = true;}
else if(pass[i] == 'O') {pass[i] = 'o'; is_Changed = true;}
}
} int main()
{
int N;
int i;
string nam,pas;
int count = 0;
cin>>N;
User *u = new User[N];
for(i=0; i<N; i++)
{
cin>>nam>>pas;
u[i].set(nam,pas);
u[i].check();
if(u[i].is_Changed) { count++; }
}
if(count == 0)
{
if(N != 1)
cout<<"There are "<<N<<" accounts and no account is modified"<<endl;
else
cout<<"There is 1 account and no account is modified"<<endl;
} else
{
cout<<count<<endl;
for(i=0; i<N; i++)
if(u[i].is_Changed) u[i].show();
}
return 0;
}

【PAT】1035. Password (20)的更多相关文章

  1. PAT 甲级 1035 Password (20 分)(简单题)

    1035 Password (20 分)   To prepare for PAT, the judge sometimes has to generate random passwords for ...

  2. 【PAT甲级】1035 Password (20 分)

    题意: 输入一个正整数N(<=1000),接着输入N行数据,每行包括一个ID和一个密码,长度不超过10的字符串,如果有歧义字符就将其修改.输出修改过多少组密码并按输入顺序输出ID和修改后的密码, ...

  3. PAT甲级——1035 Password (20分)

    To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem ...

  4. PAT Advanced 1035 Password (20 分)

    To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem ...

  5. PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642 题目描述: To prepare for PAT, the judge someti ...

  6. 【AIX】3004-314 Password was recently used and is not valid for reuse

    [AIX]3004-314 Password was recently used and is not valid for reuse   一.1  BLOG文档结构图     一.2  前言部分   ...

  7. PAT 甲级 1035 Password (20 分)

    1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for th ...

  8. 【HDU2825】Wireless Password (AC自动机+状压DP)

    Wireless Password Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u De ...

  9. 【PAT】B1075 链表元素分类(25 分)

    这道题算有点难,心目中理想的难度. 不能前怕狼后怕虎,一会担心超时,一会又担心内存过大,直接撸 将三部分分别保存到vector 有意思的在于输出 分别输出第一个的add和num 中间输出nextadd ...

随机推荐

  1. java中可以出现的中文乱码的集中解决

    从学习javaweb开始就会经常遇到中文乱码,今天就做以下记录: 1. 要避免项目中遇到乱码,首先就是在搭建项目的设置工作空间的字符编码,若是多人开发,就更应该做到统一,在eclipse中选择widn ...

  2. 【转载】CentOS日志系统组成详解

    日志系统有三部分组成:一.使用什么工具记录系统产生的日志信息?      syslog服务脚本管理的两个进程: syslogd.klogd 来记录系统产生的日志信息:      klogd     进 ...

  3. 修改linux文件权限命令:chmod 【转载】

    Linux系统中的每个文件和目录都有访问许可权限,用它来确定谁可以通过何种方式对文件和目录进行访问和操作. chmod  命令可以改变所有子目录的权限,下面有2种方法 改变一个文件的权限: chmod ...

  4. django-template-loader

    当在settings.py中设置了如下 TEMPLATE_LOADERS=( 'django.template.loaders.filesystem.Loader', 'django.template ...

  5. handler.postDelayed()和timerTask

    public static void scrollToListviewTop(final XListView listView) { listView.smoothScrollToPosition(0 ...

  6. Deamon Thread 讲解

    The daemon thread's life cycle is same with the life cycle of the application which starts this daem ...

  7. github上的QT源码,必要的时候还是应该看一下,仅凭猜测很容易出错

    QCoreApplication::processEvents 他处理的时候拿的是current不是qAppqApp的话,才是和主线程密切相关的 一直觉得QT源码复杂,有点怕,所以没怎么看 我也看不懂 ...

  8. 网盘大全, 邮箱大全 good

    网盘推荐 115网盘 注册 百度网盘 注册 微云 注册 360云盘 注册 金山快盘 注册 新浪微盘 注册 和彩云 注册 酷盘 注册 OneDrive 外链 BOX 注册 Dropbox 注册 国内网盘 ...

  9. 【从零开始,从内核驱动驱动到用户空间调用】编写第一个linux驱动,通过端口访问I/O寄存器。

    目的: 通过I/O端口方式访问RTC的秒寄存器: 由于本人从来没看过linux方面的书籍,也只是会在终端用些常用的命令而已,这次老大叫我学着通过I/O端口方式直接去读写寄存器.于是我在google中搜 ...

  10. android——使用pull解析xml文件

    1.persons.xml 将persons.xml文件放到src目录下.其代码如下: <?xml version='1.0' encoding='UTF-8' standalone='yes' ...